H.2 Euler-Maclaur in Sum Formula
16.4.1 Monatomic Ideal Gas with Gibbs Correction Factor
We give a pseudo-quantum mechanical treatment of a monatomic ideal gas, including the Gibbs correction factor. We use periodic boundary conditions and wave functions that are also eigenfunctions of the momentum operator. For a single free particle in a cubical box of volumeV, the wave function is
ψk(r)=V−1/2exp(ikãr), (16.34) which satisfies
Hψˆ k=εkψk (16.35)
withεk=¯h2k2/(2m). Here,Hˆ = ˆp2/2m, wherepˆ =(h¯/i)∇is the momentum operator. For periodic boundary conditions, the allowed values ofkare
k= 2π
V1/3[nxˆi+nyˆj+nzkˆ], (16.36)
8According to the Gibbs paradox, the additional entropy from mixing identical gases, each having the same temperature and pressure as the final mixture, turns out to be positive rather than zero, as it should be for a process that is clearly reversible. See Pathria [8, p. 22] for details.
whereˆi,ˆj,kˆare the Cartesian unit vectors andnx,ny,nzare positive and negative integers and zero. The states of a single particle of energyεlie on the surface of the sphere
n2x+n2y+n2z =2mεV2/3/h2. (16.37) The total energyE will be the sum of the energies of the individual particles because particles of an ideal gas do not, by definition, have an interaction energy. The number of quantum states forNdistinguishableparticles is therefore equal to the number of allowed solutions to9
3N
r=1
n2r =2mE V2/3/h2=R2, (16.38) where R:=(2mE V2/3/h2)1/2 is the dimensionless radius of a hypersphere in 3N- dimensional space. Allowed solutions are those for which each nr is a positive or negative integer or zero. Of course no such solutions exist unlessR2is an integer, but to circumvent this technicality we shall actually count solutions in a thin shell corresponding to energies between E − E and E, or equivalently between R − R and R, where R/R≈E/(2E).
The number of solutions to Eq. (16.38) for any radius less than or equal toRwill be equal to the volumeVRof the hypersphere given by Eq. (16.38). Obviously this volume will be proportional toR3N but the proportionality constant will depend on the dimensionality 3Nof the space. A simple derivation is given by Pathria [8, p. 504] and results in
VR= π3N/2
(3N/2)!R3N =VN(mE/2πh¯2)3N/2
(3N/2)! . (16.39)
Here,(3N/2)!should be interpreted as the gamma function(3N/2+1)in caseN is an odd integer. Since, however, 3Nis extremely large, almost all of these solutions lie near the surface of the hypersphere. In fact, the fraction of the volume of the hypersphere that lies withinRof the surface is just
F:=1−
1−R R
3N
≈1−exp(−3NR/R)≈1−exp(−3NE/2E). (16.40) The number of solutions to Eq. (16.38) in the thin shell nearEis therefore
0=FVR= [1−exp(−3NE/2E)]VR≈VR. (16.41)
9To simplify the notation, for particle number 1 we letnx=n1,ny =n2,nz=n3and for particle number 2 we letnx=n4,ny=n5,nz=n6, etc. The wave function of the whole system can be made up of products of the wave functions of the individual particles, consistent with the additivity of particle energies. Nevertheless, true quantum mechanical considerations also restrict the symmetry of the wave function under an interchange of identical particles, which is discussed in detail in Section 26.7. Here, in the spirit of treating a classical ideal gas, we omit that complication but make up for it by using the Gibbs factorN!in Eq. (16.44) to correct approximately the count of the number of microstates.
270 THERMAL PHYSICS
Thus,10
ln0=lnVR+ln[1−exp(−3NE/2E)] ≈lnVR−exp(−3NE/2E). (16.42) The second term is clearly negligible, so substitution of Eq. (16.39) and use of Stirling’s approximation gives
ln0≈lnVR∼Nln
V mE
3π¯h2N 3/2
+3
2N, (16.43)
essentially independent of any reasonable choice for E. As shown in the following example and elsewhere [8, p. 17], this result is the same as would be obtained for wave functions that vanish on the walls of the box.
One might be tempted to equate the entropy tokBln0 but that would be incorrect because ln0 is not an extensive function. The argument of the logarithm contains the ratioE/N which is intensive, but it also containsV withoutN. To get a corrected value for the number of states, we must account for the fact in observing such a system we have no way of distinguishing the particles. We follow Gibbs and divide 0by the number of indistinguishable statesN!to get the corrected number of microstates
≈0
N! =VN(mE/2π¯h2)3N/2
N!(3N/2)! . (16.44)
Thus
S=kBln≈kB[ln0−NlnN+N] (16.45) (where Stirling’s approximation has been used), which results in
S=NkBln V
N mE
3πh¯2N 3/2
+5
2NkB. (16.46)
The entropy given by Eq. (16.46) is clearly an extensive function. The temperature is given by 1/T =(∂S/∂E)V,N =(3/2)NkB/E, so in terms of the temperature we can write the entropy in the form
S=NkBln nQ/n
+5
2NkB, (16.47)
where
10Many treatments take the volume of the spherical shell to be(dVR/dR)R=VR3NR/R=VR3NE/2E and then argue that ln(VR3NE/2E)≈lnVR. That result would be obtained if the exponential in the expression forFwere expanded, which procedure is incorrect for hugeN. A more accurate expression can be obtained by settingy=(1−R/R)3Nso lny=3Nln(1−R/R)≈ −3NR/R, and then exponentiating. In fact, isolation of a system is only an idealization which is the rationale for some finiteE, notwithstanding implications of the uncertainty principle. Thus if atoms near the surface of a body were to interact weakly with its environment, one might expectE/E∼N2/3/N=N−1/3so 3NE/2E∼N2/3which is huge. After taking ln0, the additive term lnFis negligible with respect to lnVR, so ultimately one arrives at the same result as usually quoted. Our more precise analysis shows that the neglected term is much smaller than usually claimed.
nQ(T):= mkBT
2πh¯2 3/2
(16.48) is known as the quantum concentration andn:=N/V is the actual concentration. Divi- sion of0byN!is a good approximation toif the probability of multiple occupation of single particle states is negligible. Thus, Eq. (16.47) is valid provided that the actual concentrationn is small compared to the quantum concentration nQ. This will be the case at sufficiently high temperatures and low densities and will be borne out by a complete quantum mechanical analysis. From Eq. (16.46), we can calculate the chemical potential
μ= −T ∂S
∂N
E,V
=kBTln(n/nQ)=kBTln[p/(nQkBT)]. (16.49) The quantitynQkBT can be thought of as a quantum pressure. Note that(∂S/∂N)E,V = (∂S/∂N)T,V becauseE =(3/2)NkBT. On the other hand, the relationship betweenEand Tdoes not involveV for an ideal gas. Thus the pressure of an ideal gas can be computed by holding eitherEorTconstant, resulting in
p/T=∂S
∂V
N,E
=∂S
∂V
N,T
=NkB/V, (16.50)
which is the familiar ideal gas equation of state.
Example Problem 16.2. Show for an ideal gas in a box having the shape of a rectangular parallelepiped with dimensionsH,K,Lthat one obtains the same result for 0 as given by Eq. (16.41) for periodic boundary conditions and for boundary conditions for which the wave functionψ=0 on the walls of the box.
Solution 16.2. We still haveε=h¯2k2/2mand for periodic boundary conditions, k=2π
nx
H ˆi+ny K ˆj+nz
L kˆ
, (16.51)
wherenx,ny,nzare positive and negative integers and zero. Forψ=0 on the walls, the solutions are of the formψ ∝sin(kxx)sin(kyy)sin(kzz)with
k=π nx
H ˆi+ny
K ˆj+nz
L kˆ
, (16.52)
but now nx,ny,nz are only positive integers (because negative integers would only result in a change of phase, not a linearly independent eigenfunction). In the case of Eq. (16.51), nxnynz = HKL/(2π)3kxkykzso the density of states inkspace for a single particle isV/(2π)3 where the volumeV = HKL. ForN particles, the density of states is therefore
V/(2π)3N
. In the case of Eq. (16.52),nxnynz = HKL/(π)3kxkykzso the density of states inkspace for a single particle isV/(π)3, and forNparticles it is[V/(π)3]N. The volume of an entire hypersphere in 3N-dimensionalkspace with radius(2mE/¯h2)1/2is
272 THERMAL PHYSICS
Vk= π3N/2
(3N/2)!(2mE/h¯2)3N/2. (16.53) To get the corresponding number of states in the case for periodic boundary conditions, we multiplyVkby
V/(2π)3N
. But for the case ofψ =0 on the walls, only positive values ofkiare allowed, so we must first multiplyVkby(1/23)N and then by[V/(π)3]N, resulting in the same net factor[V/(2π)3]N ofVk. So in either case, the number of states (not yet corrected by the Gibbs factor) is
V (2π)3
N
Vk=VN(mE/2πh¯2)3N/2
(3N/2)! , (16.54)
the same asVRgiven by Eq. (16.39).